WAEC 2019 · Paper 2 · Q9

Coordinate geometry & circles

The curve $y = 7 - \dfrac6x$ and the line $y + 2x - 3 = 0$ intersect at two points. Find the:

(a)
coordinates of the two points;
Show the answer
$(1, 1)$ and $(-3, 9)$
(b)
equation of the perpendicular bisector of the line joining the two points.
Show the answer
$2y - x - 11 = 0$

Try it on a graph

The curve, the line, and the perpendicular bisector (purple).

Worked solution (try it first)

(a)

From the line, $y = 3 - 2x$ .
So $7 - \dfrac6x = 3 - 2x$ .
Multiply by $x$ : $7x - 6 = 3x - 2x^2$ , so $2x^2 + 4x - 6 = 0$ , that is $x^2 + 2x - 3 = 0$ .
$(x + 3)(x - 1) = 0$ : the points are $(1, 1)$ and $(-3, 9)$ .

(b)

Midpoint: $(-1, 5)$ .
Gradient of the chord: $\dfrac{9 - 1}{-3 - 1} = -2$ .
The bisector has gradient $\frac12$ : $y - 5 = \frac12(x + 1)$ , so $2y - x - 11 = 0$ .

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